Problem: Simplify and expand the following expression: $ \dfrac{4}{3q + 12}- \dfrac{4}{q - 4}- \dfrac{4q}{q^2 - 16} $
First find a common denominator by finding the least common multiple of the denominators. Try factoring the denominators. We can factor a $3$ out of denominator in the first term: $ \dfrac{4}{3q + 12} = \dfrac{4}{3(q + 4)}$ We can factor the quadratic in the third term: $ \dfrac{4q}{q^2 - 16} = \dfrac{4q}{(q + 4)(q - 4)}$ Now we have: $ \dfrac{4}{3(q + 4)}- \dfrac{4}{q - 4}- \dfrac{4q}{(q + 4)(q - 4)} $ The least common multiple of the denominators is: $ 3(q + 4)(q - 4)$ In order to get the first term over $3(q + 4)(q - 4)$ , multiply by $\dfrac{q - 4}{q - 4}$ $ \dfrac{4}{3(q + 4)} \times \dfrac{q - 4}{q - 4} = \dfrac{4(q - 4)}{3(q + 4)(q - 4)} $ In order to get the second term over $3(q + 4)(q - 4)$ , multiply by $\dfrac{3(q + 4)}{3(q + 4)}$ $ \dfrac{4}{q - 4} \times \dfrac{3(q + 4)}{3(q + 4)} = \dfrac{12(q + 4)}{3(q + 4)(q - 4)} $ In order to get the third term over $3(q + 4)(q - 4)$ , multiply by $\dfrac{3}{3}$ $ \dfrac{4q}{(q + 4)(q - 4)} \times \dfrac{3}{3} = \dfrac{12q}{3(q + 4)(q - 4)} $ Now we have: $ \dfrac{4(q - 4)}{3(q + 4)(q - 4)} - \dfrac{12(q + 4)}{3(q + 4)(q - 4)} - \dfrac{12q}{3(q + 4)(q - 4)} $ $ = \dfrac{ 4(q - 4) - 12(q + 4) - 12q} {3(q + 4)(q - 4)} $ Expand: $ = \dfrac{4q - 16 - 12q - 48 - 12q}{3q^2 - 48} $ $ = \dfrac{-20q - 64}{3q^2 - 48}$